Numerical Problems on Floatation and Metacentric Heights in Marine Engineering
Solve numerical problems related to floatation and metacentric heights in marine engineering, including calculations for metacentric height, centre of gravity, displacement, and angles of tilt. Examples involve ships, pontoons, and calculating dimensions such as breadth and height for floating structures.
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Numericals on Floatation and metacentric Heights Dr. J. Badshah Dairy Engineering Department Sanjay Gandhi Institute of Dairy Technology
Numerical on ships 1. When a load of 20 tonnes is shifted by a distance of 6 m, the ship tilts through an angle of 4 . The ship has a displacement of 5000 tonnes of sea water ( specific weight of sea water = 1025 kg/m3. If the moment of inertia of the area at the water line is 8000 m4, calculate (a) The metacentric height, and (b) the height of the centre of gravity of the vessel above the centre of Buoyancy. Solution: Volume of the displaced water V = 5000 x103/ 1025 = 4870 cu. Meter BM = I/V = 8000/4870 = 1.64 m MG = w x/W tan = 20x 6 / 5000 ( /180) x4 = 0.344 m (Taking tan = ) BG = BM MG = 1.64 0.344 = 1.296 m
Numerical on angle of heel When a weight of 588.5 KN is moved through 5 m on the deck of a ship, it causes a tilt of 0.07 radians in the ship. If the ship has a displacement of 98.1 MN , determine its metacentric height. Hence, find the angle of heel when the ship is going ahead and 4780 kN-s is being transmitted to a single propeller shaft which is rotating at 10.5 rad/s. Solution: Metacentric Height MG = wx / W tan = MG = 588.5 x 5 x 1000/98.1x106 xtan (0.07 radian) = 0.428 m In the second case, if T is the torque, then then work transmitted is wT = 4780 x 103 Therefore, Torque T = 4780 x 1000/10.5 = 4.55x 105N-m. Let be the angle of heel in second case. Tan = T/W x MG Tan = 4.55 x 105/ (98.1 x 10 6x 0.428) = 0.0108 Angle of heel = 0.621
Numerical on metacentric Heights of Pontoon A rectangular pontoon weighing 240 tonnes has a length of 20 m. The centre of gravity is 30 cm above the cetre of cross section. The metacentric height is 1.33 mm, when the angle of heel is 10 . The free board is not to be more than 0.67 m when the pontoon is vertcal. Find the breadth and height of the pontoon if it is floating in fresh water. Solution: Volume of water displaced = 240 x 10 3/ 103 = 240 m3 Let b be the breadth and D be the depth of pontoon. Therefore volume of water displaced by pontoon, 240 = b (D- 0.67) x20 Because BM = I/V = 20x b3 x1/12/240 = (1/144) b3 Height of centre of Buoyancy above base A AB= ( D- 0.67) /2 Height of centre of gravity above base A, AG = D/2 + 0.30 Therefore BG = (D/2 +0.30) (D-0.67)/2 = 0.635 m MG = BM BG, 1.33 = 1/144 b3 0.635 B = 6.58 m and 6.58 (D -0.67) 12 or D = 2.49 m
